Hydrogen gas (H2) reacts with nitrogen gas (N2) to produce ammonia (NH3)...?
2020-08-19 10:17 am
If 609.9 grams of hydrogen react with excess of nitrogen, how many grams of ammonia produced?
回答 (2)
2020-08-19 10:44 am
Molar mass of H₂ = 1.008×2 g/mol = 2.016 g/mol
Molar mass of NH₃ = (14.007 + 1.008×3) g/mol = 17.031 g/mol
Balanced equation for the reaction:
N₂ + 3H₂ → 2NH₃
Mole ratio H₂ : NH₃ = 3 : 2
Moles of H₂ reacted = (609.9 g) / (2.016 g/mol) = 302.5 mol
Moles of NH₃ produced = (302.5 mol) × (2/3) = 201.7 mol
Mass of NH₃ produced = (201.7 mol) × (17.031 g/mol) = 3435 g
====
OR:
(609.9 g H₂) × (1 mol H₂ / 2.016 g H₂) × (2 mol NH₃ / 3 mol H₂) × (17.031 g NH₃ / 1 mol NH₃)
= 3435 g NH₃
Molar mass of NH₃ = (14.007 + 1.008×3) g/mol = 17.031 g/mol
Balanced equation for the reaction:
N₂ + 3H₂ → 2NH₃
Mole ratio H₂ : NH₃ = 3 : 2
Moles of H₂ reacted = (609.9 g) / (2.016 g/mol) = 302.5 mol
Moles of NH₃ produced = (302.5 mol) × (2/3) = 201.7 mol
Mass of NH₃ produced = (201.7 mol) × (17.031 g/mol) = 3435 g
====
OR:
(609.9 g H₂) × (1 mol H₂ / 2.016 g H₂) × (2 mol NH₃ / 3 mol H₂) × (17.031 g NH₃ / 1 mol NH₃)
= 3435 g NH₃
2020-08-19 10:41 am
Balanced equation:
N2 + 3 H2 --> 2 NH3
609.9 g H2 / 2.0158 g/mol H2 X (2 mol NH3 / 3 mol H2) X 17.031 g/mol NH3 = 3435 g NH3
N2 + 3 H2 --> 2 NH3
609.9 g H2 / 2.0158 g/mol H2 X (2 mol NH3 / 3 mol H2) X 17.031 g/mol NH3 = 3435 g NH3
收錄日期: 2021-04-24 07:57:37
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20200819021756AAahQUA